On using elementary column operations C2 → C2 – 2C1 in the following matrix equation \(\left[\begin{array}{cc}1 & −3 \\ 2 & 4\end{array}\right] = \left[\begin{array}{cc}1 & −1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right]\), we have :

1
\(\left[\begin{array}{cc}1 & −5 \\ 0 & 4\end{array}\right] = \left[\begin{array}{cc}1 & −1 \\ −2 & 2\end{array}\right]\left[\begin{array}{cc}3 & −5 \\ 2 & 0\end{array}\right]\)
2
\(\left[\begin{array}{cc}1 & −5 \\ 0 & 4\end{array}\right] = \left[\begin{array}{cc}1 & −1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & −5 \\ −0 & 2\end{array}\right]\)
3
\(\left[\begin{array}{cc}1 & −5 \\ 2 & 0\end{array}\right] = \left[\begin{array}{cc}1 & −3 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ −2 & 4\end{array}\right]\)
4
\(\left[\begin{array}{cc}1 & −5 \\ 2 & 0\end{array}\right] = \left[\begin{array}{cc}1 & −1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & −5 \\ 2 & 0\end{array}\right]\)

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